Farm Power · Farm Power · 2 marks · Numerical answer
A tractor engine rejects0.58 kW to cooling per kW brake power. Required water flow is0.16 L/s per kW heat rejected to maintain6°C water drop across the radiator. For45 kW brake power, find radiator flow in L/s (two decimals).
Cooling heat rejection is0.58×45=26.1 kW. Multiply by the given flow requirement:26.1×0.16=4.176 L/s, approximately4.18. The stated proportional flow requirement already incorporates the cooling-temperature specification; do not divide by6 again.
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